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The sums of fourth power numbers

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The sum of three numbers is 6, the sum of their squares is 8, and the sum of their cubes is 5. What is the sum of their fourth powers?

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Let the numbers be a, b, and c. Then we have
a + b + c = 6
a2 + b2 + c2 = 8
a3 + b3 + c3 = 5
find the monic cubic equation
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Interesting result!


I don't solve cubics. I am retired, and I don't have to do things like that.

Or Bayes formulas, at least until I learn to understand them.

Newton-Raphson, on the other hand is a convenient and general

sledge-hammer tool, and it's fairly simple to program.

I did a manual search that convinced me the numbers

are not integers, and they are not (all) real.

NR tells me that are of the form

a = real
b = x + iy
c = x - iy


With values of

a = 2.67770427706
x = 1.66114786147
y = 1.53116371279


These values gave 6 8 and 5 for sums squares and cubes, within 10-8.

The sum of their 4th powers? Zero!

I found an alternative and interesting triplet that someone might want to solve.

It may be quite easy because [white spoiler] a, x and y are all integers [/white spoiler].

Hint: to see the spoiler, just select it.

a + b + c = 4
a2+ b2+ c2 = 4
a3+ b3+ c3 = 4
a4+ b4+ c4 = ___?

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Let the numbers be a, b, and c. Then we have
a + b + c = 6
a2 + b2 + c2 = 8
a3 + b3 + c3 = 5
find the monic cubic equation
a3 - a2 - a + 1 + b3 - b2 - b + 1 + c3 - c2 - c + 1 = 5 - 8 - 6 + 3

(a+1)(a-1)(a-1) + (b+1)(b-1)(b-1) + (c+1)(c-1)(c-1) = - 6

using excel: try for a & b

[a] [b] [c=d^(1/3)] :sum 6 ?

[=a^2] [=b^2] [=c^2] :sum 8 ?

[=a^3] [=b^3] [d=5-a^3-b^3] :sum 5 !

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Interesting result!

I don't solve cubics. I am retired, and I don't have to do things like that.

Or Bayes formulas, at least until I learn to understand them.

Newton-Raphson, on the other hand is a convenient and general

sledge-hammer tool, and it's fairly simple to program.

I did a manual search that convinced me the numbers

are not integers, and they are not (all) real.

NR tells me that are of the form

a = real

b = x + iy

c = x - iy

With values of

a = 2.67770427706

x = 1.66114786147

y = 1.53116371279

These values gave 6 8 and 5 for sums squares and cubes, within 10-8.

The sum of their 4th powers? Zero!

I found an alternative and interesting triplet that someone might want to solve.

It may be quite easy because [white spoiler] a, x and y are all integers [/white spoiler].

Hint: to see the spoiler, just select it.

a + b + c = 4

a2+ b2+ c2 = 4

a3+ b3+ c3 = 4

a4+ b4+ c4 = ___?

It appeared to me too that a,b,c all cant be real. If we could find the value of a*b*c, then the problem could be done. Can you please provide a link for newton raphson method (some easy explanation ;))

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If cubic equation x3 − Ax2 + Bx − C = 0 has roots a, b, c, then, expanding (x − a)(x − b)(x − c), we find

A = a + b + c
B = ab + bc + ca

C = abc

Then B = ab + bc + ca = ½ [(a + b + c)2 − (a2 + b2 + c2)] = 14.
Hence a, b, c are roots of x3 − 6x2 + 14x − C = 0, and we have

a3 − 6a2 + 14a − C = 0
b3 − 6b2 + 14b − C = 0
c3 − 6c2 + 14c − C = 0

Adding, we have (a3 + b3 + c3) − 6(a2 + b2 + c2) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0.
Hence C = 41/3, and x3 − 6x2 + 14x − 41/3 = 0.

Multiplying the polynomial by x, we have x4 − 6x3 + 14x2 − 41x/3 = 0. Then

a4 − 6a3 + 14a2 − 41a/3 = 0
b4 − 6b3 + 14b2 − 41b/3 = 0
c4 − 6c3 + 14c2 − 41c/3 = 0

Adding, we have (a4 + b4 + c4) − 6(a3 + b3 + c3) + 14(a2 + b2 + c2) − 41(a + b + c)/3 = 0.
Hence a4 + b4 + c4 = 6×5 − 14×8 + (41/3)×6 = 0.

That is, the sum of the fourth powers of the numbers is 0.

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If cubic equation x3 − Ax2 + Bx − C = 0 has roots a, b, c, then, expanding (x − a)(x − b)(x − c), we find

A = a + b + c

B = ab + bc + ca

C = abc

Then B = ab + bc + ca = ½ [(a + b + c)2 − (a2 + b2 + c2)] = 14.

Hence a, b, c are roots of x3 − 6x2 + 14x − C = 0, and we have

a3 − 6a2 + 14a − C = 0

b3 − 6b2 + 14b − C = 0

c3 − 6c2 + 14c − C = 0

Adding, we have (a3 + b3 + c3) − 6(a2 + b2 + c2) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0.

Hence C = 41/3, and x3 − 6x2 + 14x − 41/3 = 0.

Multiplying the polynomial by x, we have x4 − 6x3 + 14x2 − 41x/3 = 0. Then

a4 − 6a3 + 14a2 − 41a/3 = 0

b4 − 6b3 + 14b2 − 41b/3 = 0

c4 − 6c3 + 14c2 − 41c/3 = 0

Adding, we have (a4 + b4 + c4) − 6(a3 + b3 + c3) + 14(a2 + b2 + c2) − 41(a + b + c)/3 = 0.

Hence a4 + b4 + c4 = 6×5 − 14×8 + (41/3)×6 = 0.

That is, the sum of the fourth powers of the numbers is 0.

With this method we find abc=41/3. Now if we expand (a+b+c)^4 as (a+b+c)^2*(a+b+c)^2, then we get a^4+b^4+c^4=24*abc-328, which on substituting abc becomes = 0.

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dm92 asks for info on Newton method. Several links out there, this one is OK.

http://m.youtube.com/watch?v=ER5B_YBFMJo&feature=related

Successive approximations.

You solve f(x)=0. You give an initial guess x0. Then x1 = x0 - f(x0)/f'(x0). Repeat.

In several dimensions, like this problem with real numbers a x and y, the functions are

f1 = 6-sum abc

f2 = 8-sum2 abc

f3 = 5-sum3 abc

Calculate a 3x3 matrix M of partial derivatives of f123 wrt axy

Guess axy, and repeat axy = axy - M-1 x f123

You can use delta's to approximate the derivatives if f is messy.

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Note that we did not need to actually calculate a, b, and c in order to determine the sum of their fourth powers. In fact, one of the numbers is real; the other two are complex conjugates as Bonanova points out; see below. The approximate values of the numbers are 2.67770, and 1.66115 ± 1.53116i.

p079s.gif

now in response to Bonanova:

Generalization

Using the above approach, we can show that if

a + b + c = r
a2 + b2 + c2 = s
a3 + b3 + c3 = t

then a, b, c are roots

It then follows that of x3 − rx2 + ½(r2 − s)x + (½r(3s − r2) − t)/3 = 0

a4 + b4 + c4= 4rt/3 − ½s(r2 − s) + r2(r2 − 3s)/6= (r4 − 6r2s + 3s2 + 8rt)/6.


Recurrence Relation

Let f(n) = an + bn + cn, where n is a positive integer.

Then, given the equation x3 − 6x2 + 14x − 41/3 = 0, we can multiply by xn and sum over the three roots to yield the following recurrence relation:

f(n+3) = 6f(n+2) − 14f(n+1) + (41/3)f(n).

Successive applications of this formula allow us to calculate a5 + b5 + c5, a6 + b6 + c6, and so on.

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